MITP 413: Wireless Technologies Week 4 Michael L. Honig Department of ECE Northwestern University April 2005

Why Digital Communications? 1G (analog)  2G (digital)  3G (digital) Digitized voice requires about 64 kbps, therefore the required bandwidth is >> the bandwidth of the voice signal (3—4 kHz)!

Binary Frequency-Shift Keying (FSK) Bits: 1 0 1 1 0

Quadrature Phase Shift Keying (QPSK) Bits: 00 01 10 11

Binary Phase Shift Keying (BPSK) Bits: 1 0 1 1 0 Baseband signal

Amplitude Shift Keying (4-Level ASK) Bits: 00 01 10 11 Baseband signal symbol duration

Baseband  RF Conversion Passband (RF) signal Baseband signal sin 2fct time X T fc is the carrier frequency Why not transmit the baseband signal? Power signal bandwidth is roughly 1/T Power frequency frequency  0 fc

Why Modulate?

Selection Criteria How do we decide on which modulation technique to use? Which of these criteria are especially important for wireless channels (as opposed to wired channels)?

Example: Binary vs. 4-Level ASK Rate = 1/T symbols/sec Bandwidth is roughly 1/T Hz Bandwidth efficiency = 1 bps/Hz Rate = 2/T symbols/sec Bandwidth is roughly 1/T Bandwidth efficiency = 2 bps/Hz What about power efficiency?

Noisy Baseband Signals Rate = 1/T symbols/sec Bandwidth is roughly 1/T Hz Bandwidth efficiency = 1 bps/Hz Power =A2 (amplitude squared). Rate = 2/T symbols/sec Bandwidth is roughly 1/T Bandwidth efficiency = 2 bps/Hz Power = (A2 + 9A2)/2 = 5A2 What about probability of error vs transmitted power?

Probability of Error 4-ASK BPSK 7 dB (factor of 5) Signal-to-Noise Ratio (dB)

How to Increase Bandwidth Efficiency?

The Fundamental Question Given: B Hz of bandwidth S Watts of transmitted signal power N Watts per Hz of background noise (or interference) power What is the maximum achievable data rate? (Note: depends on Pe.)

Claude Shannon (1916-2001) Father of “Information Theory” His 1948 paper “A Mathematical Theory of Communications” laid the foundations for modern communications and networking. “Shannon was the person who saw that the binary digit was the fundamental element in all of communication. That was really his discovery, and from it the whole communications revolution has sprung.” -- R. Gallager (MIT) Other contributions and interests: digital circuits, genetics, cryptography, investing, chess-playing computer, roulette prediction, maze-solving, unicycle designs, juggling

Shannon’s Channel Coding Theorem (1948) noise Information Source Encoder Channel Decoder bits input x(t) output y(t) Estimated bits Information rate: R bits/second Channel capacity: C bits/second R < C  There exists an encoder/decoder combination that achieves arbitrarily low error probability. R > C  The error probability cannot be made small.

Shannon Capacity Channel capacity: C = B log(1+S/N) bits/second noise Information Source Encoder Channel Decoder bits input x(t) output y(t) Estimated bits Channel capacity: C = B log(1+S/N) bits/second B= Bandwidth, S= Signal Power, N= Noise Power No fading

Observations “There exists” does not address complexity issues. As the rate approaches Shannon capacity, to achieve small error rates, the transmitter and (especially) the receiver are required to do more and more computations. The theorem does not say anything about delay. To achieve Shannon capacity the length of the transmitted code words must tend to infinity! The previous formula does not apply with fading, multipath, frequency-selective attenuation. It has taken communications engineers more than 50 years to find practical coding and decoding techniques, which can achieve information rates close to the Shannon capacity.

Example: GSM Bandwidth = 200 kHz, S/I = 9 dB = 7.943 C = 200,000 x log(8.943) ≈ 632 kbps This is what would be achievable in the absence of fading, multipath, etc. Currently, the rate is about 270 kbps. Is the Shannon formula relevant?

Data Rates for Deep Space Applications Mariner: 1969 (Mars) Pioneer 10/11: 1972/3 (Jupiter/Saturn fly-by) Voyager: 1977 (Jupiter and Saturn) Planetary Standard: 1980’s (military satellite) BVD: “Big Viterbi Decoder” Galileo: 1992 (Jupiter) (uses BVD) Turbo Code: 1993

TYPES ALL PSK (PSK Type Summary Slide)

1000 Symbols Per Second x 1 Bit Per Symbol = 1000 Bits Per Second BPSK Bitream 0 1000 Bits Per Second 1000 Symbols (Phases) Per Second Now in an actual radio system, the modulator may transmit thousands of symbols each second. 1000 Symbols Per Second x 1 Bit Per Symbol = 1000 Bits Per Second

1000 Symbols Per Second x 1 Bit Per Symbol = 1000 Bits Per Second BPSK Bitream 1 1000 Bits Per Second 1000 Symbols (Phases) Per Second For example, if the modulator transmits 1000 symbols, or phases, each second, then with Binary PSK, since each symbol represents 1 bit, the data rate is 1000 bits per second. 1000 Symbols Per Second x 1 Bit Per Symbol = 1000 Bits Per Second

1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second QPSK Bitream 00 2000 Bits Per Second 1000 Symbols (Phases) Per Second But if we use 4-PSK, or QPSK, then each symbol represents 2 bits, and the data rate is 2000 bits per second. 1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second

1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second QPSK Bitream 10 2000 Bits Per Second 1000 Symbols (Phases) Per Second (3-bit bitstream animation continues at random) 1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second

1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second QPSK Bitream 01 2000 Bits Per Second 1000 Symbols (Phases) Per Second (2-bit bitstream animation continues at random) 1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second

1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second QPSK Bitream 11 2000 Bits Per Second 1000 Symbols (Phases) Per Second (2-bit bitstream animation continues at random) 1000 Symbols Per Second x 2 Bits Per Symbol = 2000 Bits Per Second

1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second 8PSK Bitstream 000 3000 Bits Per Second 1000 Symbols (Phases) Per Second And if we use 8-PSK, which conveys 3 bits for each symbol, then the data rate would be 3000 bits per second. 1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second

1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second 8PSK Bitstream 001 3000 Bits Per Second 1000 Symbols (Phases) Per Second (3-bit bitstream animation continues at random) 1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second

1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second 8PSK Bitstream 101 3000 Bits Per Second 1000 Symbols (Phases) Per Second (3-bit bitstream animation continues at random) 1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second

1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second 8PSK Bitstream 110 3000 Bits Per Second 1000 Symbols (Phases) Per Second (3-bit bitstream animation continues at random) 1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second

1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second 8PSK Bitstream 111 3000 Bits Per Second 1000 Symbols (Phases) Per Second (3-bit bitstream animation conlcudes) 1000 Symbols Per Second x 3 Bits Per Symbol = 3000 Bits Per Second

Colored Balls 2 The width of the funnel tube is analogous to the available bandwidth. Now the narrower the funnel, the longer it takes to get the jellybeans through the funnel. So, the less bandwidth we have, the longer it takes to transmit our symbols.

Binary Phase Shift Keying (BPSK) Bits: 1 0 1 1 0 Baseband signal

Minimum Bandwidth (Nyquist) Pulse Shape This pulse has the minimum bandwidth for a given symbol rate. Given bandwidth B, the maximum symbol rate without intersymbol interference (ISI) is B, the “Nyquist rate”.

Pulse Width vs. Bandwidth Power signal pulse bandwidth B = 1/T Narrowband frequency time 2T signal pulse Power bandwidth B = 1/T Wideband time frequency 2T

Shifted Nyquist Pulses Bits: 1 1 0 0 1 0

Baseband Waveform (Nyquist Signaling) . . . Bits: 1 1 0 0 1 0 . . .

Baseband  RF Conversion Passband (RF) signal Baseband signal sin 2fct X T time fc is the carrier frequency Why not transmit the baseband signal? Power signal bandwidth is roughly 1/T Power frequency  0 fc frequency

Passband Signal with Different Carrier Frequencies

Pulse Width vs. Bandwidth time Perfect synchronization T time Offset causes severe ISI!

Raised Cosine Pulses Minimum BW frequency time

Raised Cosine Pulses frequency time Minimum BW 50% excess BW 100% excess BW frequency time Excess bandwidth= (Total bandwidth – Nyquist bandwidth)/Nyquist bandwidth

Circle Vocoder And for voice communications, this data rate depends on the data rate generated by the vocoder.

Blocks Transition (Transitional Slide)

Blocks Stretched 8000 Bits Per Second The vocoder selected by the Hughes engineers used the same technology that was selected for the North American digital cell phone standard IS54. That vocoder generates 8000 bits per second. But we actually need to transmit significantly more than this because...

Additional Bits Additional Bits Error Correction Control Information ... we need to insert additional bits to guard against errors and to provide overhead control information. The control information tells the transmitter and receiver operational information like which ground station to use, how much power to use, the identity of the person using the service, billing information, and so forth. So, in reality, the system needs to support a data rate significantly larger than 8,000 bits per second – something in the range of approximately 12000 bits per second. Channel Ground Station Power Identity Billing

6000 Hz Available Bandwidth In our case, the FCC has given us 6000 Hz. We can estimate the maximum number of symbols, which we can transmit every second, by applying one of the basic principles of digital communications, sometimes referred to as Nyquist's Theorem. Nyquist's Theorem says that, roughly speaking, we can count on being able to send about 4000 symbols per second over a 6000 Hz channel. Nyquist’s Theorem: Can transmit 4000 symbols per second through a 6000 Hz channel

Colored Balls 2 5000 symbols per second would be quite difficult, and 4000 symbols per second is relatively easy. Nyquist’s Theorem: Can transmit 4000 symbols per second through a 6000 Hz channel

4000 bps < 8000 bps (Vocoder rate) BPSK 4000 < 8000 BPSK: 1 Bit Per Symbol 4000 Bits Per Second (bps) 4000 bps < 8000 bps (Vocoder rate) BPSK So, let's assume that we are sending 4000 symbols per second. If we use Binary PSK, then we are transmitting 1 bit for each symbol, so the data rate in that case is 4000 bits per second. This is far below the 8000 bits per second generated by the vocoder, which means that we cannot use BPSK to transmit our voice signal.

QPSK 8000 BPS QPSK: 2 Bits Per Symbol 2 X 4000 = 8000 Bits Per Second (bps) 8000 bps = Vocoder rate Need more bits for error correction and control! QPSK So now let's consider 4-PSK, or QPSK. In that case we are transmitting 2 bits per symbol, and with 4000 symbols per second, that gives a data rate of 2 times 4000, or 8000 bits per second. Recall that we need more than this, because on top of the vocoder rate of 8000 bits per second, we need to add additional bits to correct errors, and for control. This means that we cannot use QPSK either.

8PSK 12000 BPS 8PSK: 3 Bits Per Symbol 3 X 4000 = 12,000 Bits Per Second (bps) 8000 bps + 4000 bps Vocoder rate + Error Correction and Control 8PSK Moving then to 8-PSK, this gives us 3 bits per symbol, and with 4000 symbols per second, that gives a data rate of 3 times 4000, or 12,000 bits per second. This will give us the 8000 bits per second from the vocoder, plus 4000 bits per second for correcting and detecting errors, and for control. For this type of application, an additional 4000 bits per second beyond the vocoder rate gives us a reasonable margin for errors and control, so that we conclude that a voice service over this 6000 Hz channel can be supported with 8-PSK.

16 phases: 4 Bits Per Symbol 4 x 4000 = 16,000 bps 16 PSK 16000 BPS 16 phases: 4 Bits Per Symbol 4 x 4000 = 16,000 bps More than enough for vocoder rate + overhead But couldn't we also transmit more than 8 phases, or equivalently, more than 3 bits for every symbol? For example, suppose that we transmit 4 bits per symbol. This would give us a bit rate of 4 times 4000, or 16,000 bits per second, which is more than we need to support a voice service with our 8000 bit per second vocoder. We could do this, but it would make the system more complicated.

MOTION BLUR As explained in the last video, the problem with increasing the number of bits per symbol is that we have to increase the number of phases or symbols to transmit, and these become harder to distinguish at the receiver. Namely, 4 bits per symbol means that we have to choose from among 16 possible phases, and to avoid confusing these symbols at the receiver, we need to transmit with more power.

90˚ QPSK w/Bit Labels 0˚ 180˚ 270˚ Because we can choose from one of four phases, each phase can be used to represent two bits, instead of one as before. This is shown in the figure by labeling each phase with two bits. Namely, zero phase corresponds to transmitting 00, shown in green, 90 degrees corresponds to transmitting 11, shown in purple, 180 degrees corresponds to 10, shown in orange, and 270 degrees corresponds to 01, shown in red. (Bit labels fade-in one at a time)

90˚ 8PSK 45˚ 135˚ 0˚ 180˚ 225˚ 270˚ 315˚ Notice that the angle of each flag position again corresponds directly to the starting phase of the radio wave. (Each pair of colored dots connected by a line wipe down successively)

QPSK Signal Constellation amplitude = 1 x 1 x x What happens if we rotate the points? x

In-Phase/Quadrature Components x (a,b)  a sin 2fct + b cos 2fct 1 x x b is the “in-phase” signal component a is the “quadrature” signal component x

In-Phase/Quadrature Components x x 1 x x

Example Constellations quadrature QPSK BPSK in-phase x x x x x x 16-QAM quadrature 8-PSK x x x x For the 16-QAM signal constellation, what signal does a particular point represent? x x x x x x x x x x x x x x x in-phase x x x x x

Quadrature Modulation in-phase signal even bits Baseband Signal X Split: Even/Odd source bits transmitted (RF) signal + Baseband Signal X odd bits quadrature signal

Modulation for Fading Channels Problems: 1. Amplitude variations (shadowing, distance, multipath) 2. Phase variations 3. Frequency variations (Doppler) Solution to 1: Avoid amplitude modulation Power control Solution to 2 & 3: Avoid phase modulation (use FSK) “Noncoherent” demodulation: does not use phase reference Differential coding/decoding “Coherent” demodulation: Estimate phase shifts caused by channel. Increase data rate/Doppler shift ratio

Binary Frequency-Shift Keying (FSK) Bits: 1 0 1 1 0

Minimum Shift Keying (MSK) Bits: 1 0 1 1 0 Frequencies differ by ½ cycle Used in GSM

Binary Differential Modulation (i+1)st bit = 0: 0o phase shift What is the drawback of differential modulation? waveform for ith symbol (i+1)st bit = 1: 180o phase shift

Example: DQPSK x x x x x x x x constellation for ith symbol bits: 00 x 10 x x 01 x x x 11 x x constellation for ith symbol constellation for (i+1)st symbol Used in IS-136

Coherent Phase Modulation Receiver estimates phase offset More complicated than noncoherent (e.g., differential) modulation. Receiver requires a pilot signal. Transmit known symbols, measure phase. Pilot symbols are overhead (not information bits).

Probability of Error

Probability of Error with Fading

Orthogonal Frequency Division Multiplexing (OFDM) Modulate Carrier f1 substream 1 Split into M substreams Modulate Carrier f2 substream 2 source bits substream M OFDM Signal + Modulate Carrier fM

OFDM Spectrum … … f1 f2 f3 f4 f5 f6  0 Total available bandwidth Data spectrum for a single carrier Power … … f1 f2 f3 f4 f5 f6 frequency  0 subchannels M “subcarriers, or subchannels, or tones” “Orthogonal” subcarriers  no cross-channel interference.

OFDM Example: 802.11a 20 MHz bandwidth, M=64 (48 for data payload) Subchannel bandwidth = 20 MHz / 64 = 312.5 kHz Symbol rate / subchannel = 250 kilosymbols/sec Total symbol rate = 64 x 250 x 103 = 16 Msymbols/sec Bit rate? 16 QAM/subchannel  4 bits/symbol x 250 x 103 = 1 Mbps/subchannel, or 64 Mbps total 64 QAM/subchannel  6 bits/symbol x 250 x 10^3 = 1.5 Mbps/subchannel, or 96 Mbps total Includes overhead (synchronization, error correction, control) Actual data rate: 36 / 54 Mbps

Why OFDM? Single-carrier transmission also possible: 250 x 10^3 symbols/sec in 312.5 kHz means 16 Msymbols/sec would be transmitted in 20 MHz.